Ta có : \(10^{99}:9=k\left(dư1\right)\Rightarrow10^{99}=9k+1\left(k\in N^{^{\cdot}}\right)\)
\(2^3:9=h\left(dư8\right)\Rightarrow2^3=9h+8\left(h\in N\right)\)
\(\Rightarrow10^{99}+2^3=9k+1+9h+8=9k+9h+9=9\left(k+h+1\right)⋮9\Rightarrow\left(đccm\right)\)